Expected value of random exponential sum

Question

Let for $n\le X$, let $\varepsilon_n$ be either $-1$ or $1$ with probability $1/2$ (where all the $\varepsilon_n$ are independent). What would be the expected value of $$\int_0^1 \left\lvert\sum_{n\le X} \varepsilon_n e(n\alpha)\right\rvert d\alpha?$$

Answer 1

While it seems to be quite hard to calculate the expectation exactly, it is not too hard to obtain the order of magnitude in terms of $X$, by using Khinchin's inequality, which states the following:

There exist positive functions $A(p),B(p)$ such that whenever $a_1,\dots,a_m$ are complex numbers, one has, for $p>0$:

$$A(p)\left(\sum_{n=1}^m|a_n|^2\right)^{1/2}\leq\left(\mathbb{E}\left|\sum_{n=1}^ma_n\varepsilon_n\right|^p\right)^{1/p}\leq B(p)\left(\sum_{n=1}^m|a_n|^2\right)^{1/2}$$

In particular, for $p=1$: $$\mathbb{E}\int_0^1\left|\sum_{n\leq m}\varepsilon_ne^{2\pi i n\alpha}\right|\,d\alpha\approx\int_0^1\left(\sum_{n\leq m}|e^{2\pi i n\alpha}|^2\right)^{1/2}\,d\alpha=\sqrt{m}$$ Since it is known that $A(1)=\frac{1}{\sqrt{2}}, B(1)=1$, you can say that

The expected value is between $\sqrt{\frac{X}{2}}$ and $\sqrt{X}$.